Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x+3y &= -3 \\ 5x-3y &= -3\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $5x = 3y-3$ Divide both sides by $5$ to isolate $x$ $x = {\dfrac{3}{5}y - \dfrac{3}{5}}$ Substitute this expression for $x$ in the first equation. $6({\dfrac{3}{5}y - \dfrac{3}{5}}) + 3y = -3$ $\dfrac{18}{5}y - \dfrac{18}{5} + 3y = -3$ Simplify by combining terms, then solve for $y$ $\dfrac{33}{5}y - \dfrac{18}{5} = -3$ $\dfrac{33}{5}y = \dfrac{3}{5}$ $y = \dfrac{1}{11}$ Substitute $\dfrac{1}{11}$ for $y$ in the top equation. $6x+3( \dfrac{1}{11}) = -3$ $6x+\dfrac{3}{11} = -3$ $6x = -\dfrac{36}{11}$ $x = -\dfrac{6}{11}$ The solution is $\enspace x = -\dfrac{6}{11}, \enspace y = \dfrac{1}{11}$.